On some combinatorial problems in metric spaces of bounded doubling dimension


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1 On some combinatorial problems in metric spaces of bounded doubling dimension Michiel Smid February 5, 2010 Abstract A metric space has doubling dimension d if for every ρ > 0, every ball of radius ρ can be covered by at most 2 d balls of radius ρ/2. This generalizes the Euclidean dimension, because the doubling dimension of Euclidean space R d is proportional to d. The following results are shown, for any d 1 and any metric space of size n and doubling dimension d: First, the maximum number of diametral pairs is Θ(n 2 ). Second, if d = 1, the maximum possible weights of the minimum spanning tree and the allnearest neighbors graph are Θ(R log n) and Θ(R), respectively, where R is the minimum radius of any ball containing all elements of the metric space. Finally, if d > 1, the maximum possible weights of both the minimum spanning tree and the allnearest neighbors graph are Θ(Rn 1 1/d ). These results show that, for 1 d 3, metric spaces of doubling dimension d behave differently than their Euclidean counterparts. 1 Introduction Let S be a set of n points in R d, where d 1 is a constant. Let G be a graph with vertex set S, in which the weight of any edge (p, q) is defined to be the Euclidean distance pq between p and q. The weight wt(g) of G is defined to be the sum of the weights of the edges in G. School of Computer Science, Carleton University, Ottawa, Canada. This work was supported by the Natural Sciences and Engineering Research Council of Canada (NSERC). 1
2 We are interested in the maximum values (over all sets of n points) of the number of diametral pairs, the weight of the minimum spanning tree, and the weight of the allnearest neighbors graph. Diametral pairs: The diameter of S is defined to be the maximum distance between any pair of points in S. Two points p and q in S form a diametral pair, if pq is equal to the diameter of S. Let DP(S) be the number of diametral pairs in the set S. Obviously, DP(S) = 1 if d = 1. If d = 2, the maximum possible value of DP(S) is n, whereas if d = 3, the maximum possible value is 2n 2. If d 4, the value of DP(S) can be as large as Θ(n 2 ). For proofs of these claims, see Chapter 13 in Pach and Agarwal [5]. Minimum spanning tree: If S is contained in the ddimensional cube [0, L] d, then the weight wt(mst (S)) of its minimum spanning tree MST (S) is O(n 1 1/d L). This result was first shown by Few [2]. Subsequently, Smith [7], and Steele and Snyder [8] gave tighter analyses in terms of the constant factor in the BigOh bound. For the set S consisting of the n vertices on an n 1/d... n 1/d grid with cells having sides of length n 1/d L, we have wt(mst (S)) = Ω(n 1 1/d L). Thus, the above upper bound on wt(mst (S)) is tight. Allnearest neighbors graph: For any point p in S, let NN S (p) denote a nearest neighbor of p in S. That is, NN S (p) is a point q in S \ {p} for which the Euclidean distance pq is minimum. The allnearest neighbors graph ANN (S) is defined to be the directed graph with vertex set S and edge set {(p, NN S (p)) : p S}. Assume again that the point set S is contained in the cube [0, L] d. Since ANN (S) is a subgraph of MST (S), we have wt(ann (S)) = O(n 1 1/d L). The grid example given above shows that this upper bound is tight. All results mentioned above are valid in the ddimensional Euclidean space. A natural question to ask is whether or not similar results hold in an arbitrary metric space. Let S be a finite set of elements, called points, that has a distance function associated with it. This function assigns to any two points p and q in S a real number pq, which is called the distance between p and q. The set S is called a metric space, if for all points p, q, and r in S, 2
3 1. pp = 0, 2. if p q, then pq > 0, 3. pq = qp, and 4. pq pr + rq. The fourth property is called the triangle inequality. If c is a point in S and R > 0 is a real number, then the ball with center c and radius R is defined to be the set {p S : cp R}. The radius of a subset S of S is defined to be the minimum radius of any ball that contains S. Observe that the center of this ball is not necessarily an element of S. For example, the set S of n points, in which each pair of distinct points has distance R, is a metric space of radius R. Obviously, we have DP(S) = ( n 2), wt(mst (S)) = (n 1)R, and wt(ann (S)) = nr. Moreover, these are the largest possible values in terms of n and R. Thus, in order to get nontrivial results, we have to consider restricted classes of metric spaces. In this paper, we consider metric spaces of bounded doubling dimension, which were introduced by Assouad [1]; see also the book by Heinonen [4]. Definition 1 Let S be a finite metric space and let λ be the smallest integer such that the following is true: For each real number ρ > 0, every ball in S of radius ρ can be covered by at most λ balls of radius ρ/2. The doubling dimension of the metric space S is defined to be log λ. Observe that every metric space of n points has doubling dimension at most log n. Furthermore, it is not difficult to prove that the doubling dimension of the Euclidean metric in R d is proportional to d. In this paper, we prove the following results, for any metric space S consisting of n points and having doubling dimension d: 1. If d 1, then DP(S) is Θ(n 2 ) in the worst case. Thus, the maximum possible value of DP(S) is very different from the corresponding value in Euclidean space R d for 1 d 3. In fact, we show that, for d = 1, the maximum possible value of DP(S) is n 2 /4. For d 4, the bound coincides with the corresponding bound for R d. 3
4 2. If d = 1, then wt(mst (S)) is Θ(R log n) in the worst case, whereas wt(ann (S)) is Θ(R) in the worst case. Thus, metric spaces of doubling dimension 1 behave differently than their onedimensional Euclidean counterparts. 3. If d > 1, then both wt(mst (S)) and wt(ann (S)) are Θ(Rn 1 1/d ) in the worst case. Thus, these bounds coincide with the corresponding bounds for Euclidean space R d. Related work: In recent years, metric spaces of bounded doubling dimension have received a lot of attention in the algorithms literature; see HarPeled and Mendel [3] and the references given in that paper. In [3], it was already observed that these spaces are different from their Euclidean counterparts: In Euclidean space R d, the allnearest neighbors graph can be computed in O(n log n) time, whereas the worstcase running time becomes Θ(n 2 ) even if the doubling dimension is equal to 1. Smid [6] defined the following metric space, based on an earlier example in [3]: Let S = {p 1, p 2,..., p n } and let 0 < ɛ < 1 be a real number. For each i and j with 1 i j n, we define 0 if i = j, p i p j = p j p i = 4 j if i = 1 and j > 1, 4 j + ɛ otherwise. Then S is a metric space of doubling dimension 1 and radius R = 4 n + ɛ. We have NN S (p 1 ) = p 2 and, for each j with 2 j n, NN S (p j ) = p 1. Thus, both ANN (S) and MST (S) are equal to the stargraph consisting of all edges (p 1, p j ), 2 j n. (In contrast, it is well known that in Euclidean space R d, the maximum degrees of both these graphs are upper bounded by a function that depends only on d.) Both wt(ann (S)) and wt(mst (S)) are close to 4R/3. The rest of this paper is organized as follows. In Section 2, we prove the upper bounds on DP(S), wt(ann (S)), and wt(mst (S)), for any finite metric space of doubling dimension d. The proofs will be by induction on the number of points in S. Some care has to be taken, because the doubling dimension d of a subset of S may be larger than d. We show that d is always at most 2d. 4
5 In Section 3, we prove that the upper bounds in Section 2 are tight. The lower bounds for DP(S), wt(ann (S)), and wt(mst (S)) are in fact obtained by one single class of metric spaces. We conclude in Section 4 by presenting some more results that show that metric spaces of doubling dimension 1 are very different than the Euclidean space R 1. 2 The upper bounds We start by showing that DP(S) n 2 /4 for every metric space S of size n and doubling dimension 1. The proof uses the following lemma which states that in any isosceles triangle, the base is shorter than the two sides of equal length. Lemma 1 Let S be a finite metric space of doubling dimension 1, and let p, q, and r be three pairwise distinct points in S such that pq = pr. Then qr < pq. Proof. The proof is by contradiction. Let p, q, and r be a smallest triple of points for which the lemma does not hold. That is, let l be the smallest real number such that there exist three points p, q, and r in S such that l = pq = pr and qr l. Let B be the ball with center p and radius l. Observe that B contains the points p, q, and r. Since S has doubling dimension 1, we can cover B by two balls B 1 and B 2 centered at c 1 and c 2, respectively, and having radius l/2. We may assume without loss of generality that p is contained in B 1. Let l = qr. If both q and r are contained in B 2, then l = qr qc 2 + c 2 r l/2 + l/2 = l l. It follows that c 2 q = c 2 r = l/2. Since qr = l l/2, the triple c 2, q, and r forms a smaller counterexample to the lemma, which is a contradiction. Thus, q and r are not both contained in B 2. We may assume without loss of generality that q is contained in B 1. We have l = pq pc 1 + c 1 q l/2 + l/2 = l, which implies that c 1 p = c 1 q = l/2. Since pq = l l/2, the triple c 1, p, and q forms a smaller counterexample to the lemma, which is again a contradiction. 5
6 Theorem 1 Let S be a metric space of doubling dimension 1 and consisting of n points. Then DP(S) n 2 /4. Proof. Let R be the diameter of S, let c be a point in S, and let B be the ball with center c and radius R. Observe that B contains all points of S. Cover B by two balls B 1 and B 2, both having radius R/2, and define S 1 = S B 1 and S 2 = S \ S 1. Assume that S 1 contains two points p and q such that pq = R. Then, denoting the center of B 1 by c, R = pq pc + cq R/2 + R/2 = R and, therefore, cp = cq = R/2, which contradicts Lemma 1. Thus, the diameter of S 1 is less than R. By the same argument, the diameter of S 2 is less than R. If we denote the size of S 1 by m, then it follows that DP(S) is at most m(n m). Since the function m(n m), for 0 m n, is maximized when m = n/2, the proof is complete. The proofs of the upper bounds on the weights of MST (S) and ANN (S) will be by induction on the size of the metric space S. We have to be careful, however, because the doubling dimension d of a subset S of S may be larger than the doubling d dimension of S. The reason is that the value of d is determined by only considering balls that are centered at points of S. To give an example, let n = m 2 and let S be a metric space of size n, in which the distance between any two distinct points is equal to 2. The doubling dimension d of S is equal to d = log n = 2 log m. Partition S into subsets S 1, S 2,..., S m, each consisting of m points. Define S = S {p 1, p 2,..., p m }, where the p i s are new points. For each i with 1 i m, and for each q in S i, we define p i q = qp i = 1. For any other pair of distinct points in S, their distance is defined to be 2. Observe that S is a metric space. We claim that the doubling dimension d of S is equal to d = log(m + 1). To prove this claim, let ρ > 0 be a real number and let B be a ball in S of radius ρ. If ρ 2, then B is covered by m balls of radius ρ/2, centered at the points p 1, p 2,..., p m. If 1 ρ < 2, then B contains at most m + 1 points and, therefore, we can cover B by m + 1 balls of radius ρ/2. Finally, if ρ < 1, then B is a singleton set and we can cover B by one ball of radius ρ/2. Thus, d is indeed equal to log(m + 1). 6
7 For large values of m, the ratio d /d converges (from below) to 2. The next lemma states that this ratio can never be larger than 2. Lemma 2 Let S be a finite metric space of doubling dimension d and let S be a nonempty subset of S. Then the doubling dimension of the metric space S is at most 2d. Proof. Let c be a point of S, let ρ > 0 be a real number, and let B = {p S : cp ρ}. Thus, B is a ball in the metric space S. Consider the corresponding ball B in S, i.e., B = {p S : cp ρ}. Since S has doubling dimension d, we can cover B by balls B 1, B 2,..., B k, all having radius ρ/4, where k 2 2d. For 1 i k, let c i be the center of B i. If c i S, then we define B i to be the ball in S of radius ρ/2 that is centered at c i. If c i S and B i S, then we define B i to be the ball in S of radius ρ/2 that is centered at an arbitrary point of B i S. We claim that the collection of balls {B i : 1 i k, B i S } cover the ball B. To prove this, let p be a point in B. Then p B and, thus, there is an index i such that p B i. If c i S, then p B i. Otherwise, let c i be the center of B i. Since c ip c ic i + c i p ρ/4 + ρ/4 = ρ/2, it follows that p B i. Thus, we have covered the ball B in S by at most 2 2d balls (again, in S ) of radius ρ/2. Theorem 2 Let S be a metric space consisting of n points, let d be the doubling dimension of S, and let R be the radius of S. 1. If d = 1, then wt(mst (S)) 2R log n. 2. If d > 1, then wt(mst (S)) 12Rn 1 1/d. Proof. We start by considering the case when d = 1. We will prove the following claim by induction on the size of V : Let V be a nonempty subset of S and let R V be the radius of V. Then wt(mst (V )) 2R V log V. (1) By taking V = S, this claim will prove the first part of the theorem. 7
8 If V is a singleton set, then (1) obviously holds. Assume that V contains at least two elements and, further, assume that (1) holds for all nonempty subsets of S having less than V elements. Let B be a ball of radius R V that contains V. Since the doubling dimension of S is equal to one, we can cover B by two balls B 1 and B 2, both having radius R V /2. Note that the centers of B 1 and B 2 are not necessarily points of V. Define V 1 = V B 1 and V 2 = V \ V 1. Then both V 1 and V 2 are nonempty (otherwise, the radius of V would be less than R V ). Let T be the spanning tree of V consisting of MST (V 1 ), MST (V 2 ), and one edge joining an arbitrary point of V 1 with an arbitrary point of V 2. Let R V1 and R V2 be the radius of V 1 and V 2, respectively. Then R V1 R V /2 and R V2 R V /2. Since the diameter of V is at most 2R V, and by applying the induction hypothesis, we obtain wt(t ) 2R V1 log V 1 + 2R V2 log V 2 + 2R V R V (log V 1 + log V 2 + 2). Since the function log x is convex for x > 0, and since V 1 + V 2 = V, it follows that wt(t ) R V (log( V /2) + log( V /2) + 2) = 2R V log V. Since wt(mst (V )) wt(t ), we have shown that (1) holds for V. In the rest of the proof, we consider the case when d > 1. We will prove the following claim, again by induction on the size of V : Let V be a nonempty subset of S and let R V be the radius of V. Then wt(mst (V )) 12R V V 1 1/d 12R V. (2) By taking V = S, this claim will prove the second part of the theorem. If V is a singleton set, then wt(mst (V )) = R V = 0 and (2) holds. Assume that V consists of two points p and q. Then wt(mst (V )) = pq 2R V. Thus, (2) holds if we can show that /d 12. This inequality can be rewritten as 7 2 1/d 12. The latter inequality holds, because d log 3. From now on, we assume that V contains at least three elements and (2) holds for all nonempty subsets of S having less than V elements. Let B be a ball of radius R V that contains V. Since the doubling dimension of S is equal to d, we can cover B V by balls B 1, B 2,..., B k, all having radius R V /2, where k 2 d. Observe that, since the radius of V is equal to R V, we 8
9 have k 2. For 1 i k, let V i be the set of points of V that are in the ball B i. In case a point of V is contained in more than one ball, we put it in exactly one subset V i. Thus, the sets V 1, V 2,..., V k are pairwise disjoint and their union is equal to V. We may also assume that none of the sets V i is empty. We next claim that we may assume that k 3. Indeed, if k = 2, then, since V 3, one of the sets V 1 and V 2, say V 2, contains at least two points. We partition V 2 into two disjoint nonempty subsets, which we call V 2 and V 3, and, thus, we can set k = 3. Let i be an index with 1 i k, and let R Vi be the radius of the set V i. Then R Vi R V /2. By the induction hypothesis, we have Since V i 1 1/d 1 0, it follows that wt(mst (V i )) 12R Vi ( Vi 1 1/d 1 ). wt(mst (V i )) 6R V ( Vi 1 1/d 1 ). For each i with 1 i k, choose an arbitrary point in V i, and let T be an arbitrary spanning tree of these k points. Since the diameter of V is at most 2R V, we have wt(t ) 2(k 1)R V 2kR V. Let T be the spanning tree of V consisting of T and MST (V 1 ), MST (V 2 ),..., MST (V k ). Then wt(t ) k ( 6R V Vi 1 1/d 1 ) + 2kR V i=1 k = 6R V V i 1 1/d 4kR V i=1 k 6R V V i 1 1/d 12R V. i=1 Since the function x 1 1/d is convex for x > 0, and since k i=1 V i = V, it follows that k V i 1 1/d i=1 k ( V /k) 1 1/d = k 1/d V 1 1/d 2 V 1 1/d. i=1 9
10 Thus, we have wt(t ) 12R V V 1 1/d 12R V. Since wt(mst (V )) wt(t ), we have shown that (2) holds for V. Theorem 3 Let S be a metric space consisting of n points, where n 2, let d be the doubling dimension of S, and let R be the radius of S. 1. If d = 1, then wt(ann (S)) 4R. 2. If d > 1, then wt(ann (S)) 12Rn 1 1/d. Proof. Since ANN (S) is a subgraph of MST (S), the second claim follows from Theorem 2. Assume that d = 1. The first claim will follow from the following more general claim: For any subset V of S with V 2 and having radius R V, p, NN S (p) 4R V. (3) p V The proof is by induction on the size of V. If V consists of two points a and b, then p V p, NN S(p) 2 ab 4R V, thus (3) holds. Assume that V contains at least three elements and, further, assume that (3) holds for all subsets of S having at least two and less than V elements. Let B be a ball of radius R V that contains V. We cover B by two balls B 1 and B 2, both having radius R V /2, and define V 1 = V B 1 and V 2 = V \ V 1. Observe that both V 1 and V 2 are nonempty. Let R V1 and R V2 be the radius of V 1 and V 2, respectively. Then R V1 R V /2 and R V2 R V /2. If both V 1 and V 2 contain at least two points, then, using the induction hypothesis, p, NN S (p) = p, NN S (p) + p, NN S (p) p V 1 p V 2 p V 4R V1 + 4R V2 4R V. Assume that one of V 1 and V 2, say V 1, consists of one point a. Since the set V has radius R V, we have a, NN S (a) R V. Combining this with the induction hypothesis, we obtain p, NN S (p) = a, NN S (a) + p, NN S (p) R V + 4R V2 4R V. p V 2 p V 10
11 This completes the proof. 3 The lower bounds In this section, we show that the upper bounds in Theorems 1, 2 and 3 are tight. In fact, we give an example of one class of metric spaces that achieve the upper bounds in all these theorems. This class is obtained by repeatedly applying the following transformation: Let m 2 be an integer and assume we are given a metric space S consisting of m points. Let d be the doubling dimension of S, let R be the radius of S, and assume that the diameter of S is equal to R. We define a new metric space S in the following way. For each i with 1 i 2 d, let S i be a copy of the metric space S, and let S be the union of these copies. For any two points p and q in S, if they are in the same copy S i, then the distance pq (in S) is defined to be their distance in S i. Otherwise, we define pq to be 2R. Let n be the size of S, so that n = 2 d m. The following claims can easily be verified. The triangle inequality holds for the set S and, thus, S is a metric space. Both the diameter and the radius of S are equal to R = 2R and ( 2 d DP(S) = 2 ) m 2 = 1 2 (1 12 d ) n 2. (4) Since the diameter of each set S i is equal to R, and since distances between points in S i and points in S j (for j i) are equal to 2R, we have wt(ann (S)) = 2 d wt(ann (S )). (5) By running Kruskal s minimum spanning tree algorithm on S, it follows that wt(mst (S)) = (2 d 1)R + 2 d wt(mst (S )). (6) We finally claim that the doubling dimension of S is equal to d. To prove this, let c be a point in S, let ρ > 0 be a real number, and consider the ball B = {p S : cp ρ}. First assume that ρ < 2R. Then B is completely contained in one of the sets S i. Since the doubling dimension of S i is equal to d, and since distances 11
12 in S i are the same as distances in S, it follows that B can be covered by at most 2 d balls of radius ρ/2. Now assume that ρ 2R. Since the radius of S i is equal to R, there exists a ball B i of radius R that contains S i. The union of these balls B i, for 1 i 2 d, cover the entire set S and, thus, it covers the ball B. Clearly, each such ball has radius at most ρ/2. Thus, also in this case, B can be covered by at most 2 d balls of radius ρ/2. Theorem 4 Let λ 2 be an integer, let d = log λ, let k be a positive integer, and let n = 2 kd. There exists a metric space S of size n and doubling dimension d, such that the following are true: 1. Both the radius and diameter of S are equal to R = 2 k. ( 2. DP(S) = ) /2 d n 2. In particular, if d = 1, then DP(S) = n 2 /4. 3. wt(ann (S)) = 2Rn 1 1/d. In particular, if d = 1, then wt(ann (S)) = 2R. 4. If d = 1, then wt(mst (S)) = R log n. 5. If d > 1, then wt(mst (S)) 2Rn 1 1/d. Proof. Let S be the metric space of size m = 2 d, in which the distance between any two distinct points is equal to 2. The doubling dimension of S is equal to d, and both its radius and diameter are equal to R = 2. By applying the above construction k 1 times, we obtain a metric space S of size n = 2 kd, doubling dimension d, radius R = 2 k and diameter R = 2 k, proving the first claim. The second claim follows from (4). Since wt(ann (S )) = 2m = 2 2 d, it follows from (5) that wt(ann (S)) = 2 2 kd = 2n. Since n = 2 kd and R = 2 k, we have Rn 1 1/d = n. Therefore, we have wt(ann (S)) = 2Rn 1 1/d, proving the third claim. The fourth claim follows from the recurrence in (6). Finally, the fifth claim follows from the third one, because MST (S) contains ANN (S). 4 Concluding remarks We have shown that metric spaces of low doubling dimension are very different than lowdimensional Euclidean spaces. In this final section, we present 12
13 some more consequences of Lemma 1 for metric spaces of doubling dimension 1. Lemma 3 Let S be a finite metric space of doubling dimension 1, and let p, q, and r be three points in S such that (p, q) is an edge in MST (S) and pq pr qr. Then pr 2 pq. Proof. Let l = pq, l = pr, and l = qr. Thus, we have l l l. Let B be the ball with center p and radius l. This ball contains the points p, q, and r. Cover B by balls B 1 and B 2 centered at c 1 and c 2, respectively, and having radius l /2. We may assume without loss of generality that p is contained in B 1. Assume that r is in B 1. Then l = pr pc 1 + c 1 r l /2 + l /2 = l and, therefore, c 1 p = c 1 r = l /2 and pr = l, contradicting Lemma 1. Thus, r is in B 2. If we assume that q is in B 2, then, by a similar argument, it follows that c 2 q = c 2 r = l /2 and qr = l, contradicting Lemma 1. Thus, q is in B 1. We have shown that both p and q are in B 1, and that r is in B 2. Let T = MST (S). By removing the edge (p, q) from T, we obtain two trees T 1 and T 2, where p is a node in T 1 and q is a node in T 2. We may assume without loss of generality that c 1 is a node in T 1. Observe that (c 1, q) is not an edge in T. Let T be the tree obtained from T by replacing the edge (p, q) by the edge (c 1, q). Since T is a minimum spanning tree of S, it follows that l = pq c 1 q. On the other hand, since q is in B 1, we have c 1 q l /2. It follows that l 2l. We now use Lemma 3 to prove the following claims. First, by following any (directed) path in ANN (S), edges decrease in weight by a factor of at least 2. Second, for any point in S, any two incoming edges in ANN (S) differ in weight by a factor of at least 2. Finally, any two edges in MST (S) that share a point differ in length by a factor of at least 2. Lemma 4 Let S be a finite metric space of doubling dimension 1, and let p, q, and r be three pairwise distinct points in S. 1. If q = NN S (p) and p = NN S (r), then pr 2 pq. 13
14 2. If p = NN S (q), p = NN S (r), and pq pr, then pr 2 pq. 3. If both (p, q) and (p, r) are edges in MST (S) and pq pr, then pr 2 pq. Proof. The assumptions in the first and second claims imply that (p, q) is an edge in MST (S) and pq pr qr. Therefore, these two claims follow from Lemma 3. For the third claim, observe that (q, r) is not an edge in MST (S). It follows that pq pr qr and, therefore, the claim also follows from Lemma 3. References [1] P. Assouad. Plongements lipschitziens dans R N. Bulletin de la Société Mathématique de France, 111: , [2] L. Few. The shortest path and the shortest road through n points in a region. Mathematika, 2: , [3] S. HarPeled and M. Mendel. Fast construction of nets in lowdimensional metrics and their applications. SIAM Journal on Computing, 35: , [4] J. Heinonen. Lectures on Analysis on Metric Spaces. SpringerVerlag, Berlin, [5] J. Pach and P. K. Agarwal. Combinatorial Geometry. John Wiley & Sons, New York, [6] M. Smid. The weak gap property in metric spaces of bounded doubling dimension. In Efficient Algorithms, Essays Dedicated to Kurt Mehlhorn on the Occasion of His 60th Birthday, volume 5760 of Lecture Notes in Computer Science, pages , Berlin, SpringerVerlag. [7] W. D. Smith. Studies in Computational Geometry Motivated by Mesh Generation. Ph.D. thesis, Priceton University, Princeton, NJ,
15 [8] J. M. Steele and T. L. Snyder. Worstcase growth rates of some classical problems of combinatorial optimization. SIAM Journal on Computing, 18: ,
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